Chapter 10, Problem 10.85EP

Selected information about five solutions, each at 24°C, is given in the following table. Fill in the blanks in each line in the table. The first line is already completed as an example.

Interpretation Introduction

Interpretation:

For a given solution, $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ $\left[{\text{OH}}^{\text{-}}\right]$, $\text{pH}$ and whether the solution is acidic or basic that has to be determined.

Concept Introduction:

pH:

$\text{pH}$ is a scale used to specify how acidic or basic a solution is.  It ranges from 0-14. $\text{pH}$ 7.0 is considered as neutral solution, $\text{pH}$ more than 7.00 is taken as basic solution whereas $\text{pH}$ less than 7.0 is considered as acidic solution (at 25^oC). It is the measurement of activity of free ${\text{H}}^{\text{+}}$ and ${\text{OH}}^{\text{-}}$ in solution.

  $\text{pH}\text{=}-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$

  $\text{pOH}\text{=}-\log \left[{\text{OH}}^{-}\right]$

From ionization constant of water ${\text{K}}_{\text{w}}\text{=}\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]$

Ionization constant of water at ${\text{25}}^{\text{ο}}\text{C}$ ${\text{K}}_{\text{w}}\text{=}1\times {10}^{-14}$.

Less the $\text{pH}$ more is the acidity of the solution.

Explanation of Solution

Given that,

  $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=7.2\times {10}^{-10}$

Then,

  $\begin{array}{l}{\text{K}}_{\text{w}}\text{=}\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\\ \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\text{=}{\text{1×10}}^{\text{-14}}\\ \\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{{\text{1×10}}^{\text{-14}}}{\text{7}{\text{.2×10}}^{\text{-10}}}\\ \\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\text{1}{\text{.4×10}}^{\text{-5}}\end{array}$

  $\begin{array}{l}\text{pH}\text{=}-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ \text{pH}\text{=}-\log (7.2{\text{×10}}^{\text{-10}})\\ \\ \text{pH}\text{=}7.21\end{array}$

In this solution $\left[{\text{OH}}^{\text{-}}\right]$ is greater than $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ and $\text{pH}$ more than 7.00.

Hence this is a basic solution having $\text{pH}\text{=}7.21$ and $\left[{\text{OH}}^{\text{-}}\right]\text{=}\text{1}{\text{.4×10}}^{\text{-5}}\text{M}$.

Interpretation Introduction

Interpretation:

For a given solution having given $\text{pH}$, the value of $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$, $\left[{\text{OH}}^{\text{-}}\right]$ and whether the solution is acid or basic that has to be determined.

Concept Introduction:

pH:

$\text{pH}$ is a scale used to specify how acidic or basic a solution is. It ranges from 0-14.  $\text{pH}$ 7.0 is considered as neutral solution, $\text{pH}$ more than 7.00 is taken as basic solution whereas $\text{pH}$ less than 7.0 is considered as acidic solution (at 25^oC). It is the measurement of activity of free ${\text{H}}^{\text{+}}$ and ${\text{OH}}^{\text{-}}$ in solution.

  $\text{pH}\text{=}-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$

  $\text{pOH}\text{=}-\log \left[{\text{OH}}^{-}\right]$

From ionization constant of water ${\text{K}}_{\text{w}}\text{=}\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]$

Ionization constant of water at ${\text{25}}^{\text{ο}}\text{C}$ ${\text{K}}_{\text{w}}\text{=}1\times {10}^{-14}$.

Less the $\text{pH}$ more is the acidity of the solution.

Explanation of Solution

Given that,

  $\text{pH}\text{=}5.30$

Then,

  $\begin{array}{l}\text{pH}\text{=}-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ -\log \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=5.30\\ \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]={10}^{-5.30}\\ \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=5.00\times {10}^{-6}\text{M}\end{array}$

  $\begin{array}{l}{\text{K}}_{\text{w}}\text{=}\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\\ \\ \left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\text{=}{\text{1×10}}^{\text{-14}}\\ \\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{{\text{1×10}}^{\text{-14}}}{\text{5}{\text{.0×10}}^{\text{-6}}}\\ \\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\text{2}{\text{.0×10}}^{\text{-9}}\text{M}\end{array}$

In this solution $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ is greater than $\left[{\text{OH}}^{\text{-}}\right]$ and $\text{pH}$ less than 7.00 and hence the solution is acidic.

Hence this is an acidic solution with $\left[{\text{OH}}^{\text{-}}\right]\text{=}\text{2}{\text{.0×10}}^{\text{-9}}\text{M}$ and $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=5.00\times {10}^{-6}\text{M}$.

Interpretation Introduction

Interpretation:

For a given solution having $\left[{\text{OH}}^{\text{-}}\right]$, the value of $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ and $\text{pH}$ of the solution has to be calculated along with the prediction that whether the solution is acidic or basic.

Concept Introduction:

pH:

$\text{pH}$ is a scale used to specify how acidic or basic a solution is.  It ranges from 0-14. $\text{pH}$ 7.0 is considered as neutral solution, $\text{pH}$ more than 7.00 is taken as basic solution whereas $\text{pH}$ less than 7.0 is considered as acidic solution (at 25^oC). It is the measurement of activity of free ${\text{H}}^{\text{+}}$ and ${\text{OH}}^{\text{-}}$ in solution.

  $\text{pH}\text{=}-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$

  $\text{pOH}\text{=}-\log \left[{\text{OH}}^{-}\right]$

From ionization constant of water ${\text{K}}_{\text{w}}\text{=}\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]$

Ionization constant of water at ${\text{25}}^{\text{ο}}\text{C}$ ${\text{K}}_{\text{w}}\text{=}1\times {10}^{-14}$.

Less the $\text{pH}$ more is the acidity of the solution.

Explanation of Solution

Given that,

  $\left[{\text{OH}}^{\text{-}}\right]\text{=}7.2{\text{×10}}^{\text{-10}}\text{M}$

  $\begin{array}{l}{\text{K}}_{\text{w}}\text{=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\\ \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\text{=}{\text{1×10}}^{\text{-14}}\\ \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\frac{{\text{1×10}}^{\text{-14}}}{\text{7}{\text{.2×10}}^{\text{-10}}}\\ \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\text{1}{\text{.4×10}}^{\text{-5}}\end{array}$

  $\begin{array}{l}\text{pH}\text{=}-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ \text{pH}\text{=}-\log (1.4{\text{×10}}^{\text{-5}})\\ \\ \text{pH}\text{=}4.85\end{array}$

As in this solution $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ is greater than $\left[{\text{OH}}^{\text{-}}\right]$ and $\text{pH}$ less than 7.00 and hence the solution is acidic.

Hence this is an acidic solution having $\text{pH}\text{=}4.85$ and $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\text{1}{\text{.4×10}}^{\text{-5}}\text{M}$.

Interpretation Introduction

Interpretation:

For a given $\text{pH}$, the value of $\left[{\text{OH}}^{\text{-}}\right]$, $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ and whether the solution is acid or basic that has to be determined.

Concept Introduction:

pH:

$\text{pH}$ is a scale used to specify how acidic or basic a solution is.  It ranges from 0-14. $\text{pH}$ 7.0 is considered as neutral solution, $\text{pH}$ more than 7.00 is taken as basic solution whereas $\text{pH}$ less than 7.0 is considered as acidic solution (at 25^oC). It is the measurement of activity of free ${\text{H}}^{\text{+}}$ and ${\text{OH}}^{\text{-}}$ in solution.

  $\text{pH}\text{=}-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$

  $\text{pOH}\text{=}-\log \left[{\text{OH}}^{-}\right]$

From ionization constant of water ${\text{K}}_{\text{w}}\text{=}\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]$

Ionization constant of water at ${\text{25}}^{\text{ο}}\text{C}$ ${\text{K}}_{\text{w}}\text{=}1\times {10}^{-14}$.

Less the $\text{pH}$ more is the acidity of the solution.

Explanation of Solution

Given that,

$\text{pH}\text{=}12.00$

Then,

  $\begin{array}{l}\text{pH}\text{=}-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ -\log \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=8.23\\ \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]={10}^{-8.23}\\ \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=5.8\times {10}^{-9}\end{array}$

  $\begin{array}{l}{\text{K}}_{\text{w}}\text{=}\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\\ \\ \left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]=1\times {10}^{-14}\\ \\ \left[{\text{OH}}^{\text{-}}\right]=\frac{1\times {10}^{-14}}{\text{5}{\text{.8×10}}^{\text{-9}}}\\ \\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\text{1}{\text{.7×10}}^{\text{-6}}\end{array}$

In this solution $\left[{\text{OH}}^{\text{-}}\right]$ is greater than $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ and $\text{pH}$ more than 7.00. hence this solution is basic solution.

Hence this is a basic solution having $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\text{5}{\text{.8×10}}^{\text{-9}}\text{M}$ and $\left[{\text{OH}}^{\text{-}}\right]\text{=}\text{1}{\text{.7×10}}^{\text{-6}}\text{M}$.